Proof:
()
If is an isometry, then by direct calculation we know that is a projection. Hence and is a unitary.
()
If for some projection . Then by the definition of murray von neumann equivalence, there exists an isometry which connects and . Thus and .But in this algebra, every isometry is a unitary, so , this shows that is equivalent to a proper subprojection of itself.
()
Suppose that and are two projections in such that and . Let be a partial isometry in which implements the equivalence between and . and . Put and from our assumption, so we get that . We conclude that and , but since all isometries are unitary, necessarily, proving that every projection is equivalent to some proper sub projection of itself.
() If holds, then the unit is a finite projection.
() Parse through what it means.
() Trivial.
()
Suppose that holds, and let be a left-invertible element in . find such that . Recall that if are self-adjoint elements in a unital -algebra , then and if then for each . Use these facts to obtain the inequality This shows that and the spectrum of is therefore contained in the interval , which in particular implies tat is invertible. Put , and observe that therefore is invertible by (2), and hence is invertible.
Warnings: